1. (07.03 MC) Solve -+ 2 2. 5 7 - (4 points) 35 0 24 7

What is the approximate area that is shaded red?

Mashcka [7]

To find the area that is shaded red, you find the area of the rectangle and subtract it by the area of the triangle.

Area of a rectangle:

A = l × w [ l = 11 in ; w = 6in ] Plug these numbers into the equation

A = 11 · 6

A = 66 in²

Area of a triangle:

$A=\frac{1}{2}bh$ [ b = 4 in ; h = 6 in ] Plug these #s into the equation

$A = \frac{1}{2}(4)(6)$

$A=\frac{24}{2}$

A = 12 in²

Area of rectangle - Area of triangle = Area of the shaded region

66 in² - 12 in ² = 54 in²

Your answer is B

3 0

3 years ago

Please zoom in before you start solving because there are tricks in here. What is the answer?

blondinia [14]

45 ÷ 3 = 15
21 ÷ 3 = 7
12 ÷ 3 = 4

(4 + 15) x 7 = 133

OR if you’re using bedmas/pedmas

4 + (15 x 7) = 109

7 0

3 years ago

When one power is exponentiated by an exponent, what is done with the exponents?

Mama L [17]

I’m pretty sure they are multiplied, if I understand your wording

3 0

2 years ago

Help ASAP. Thank you! Find the area of the shaded region. 80° 5 cm A=[?] cm2

pashok25 [27]

Answer:

Step-by-step explanation:

$\text{The area shaded outside of the triangle is}\\ \\ A=\frac{\pi r^2\theta}{360},\ r=5,\ \theta =80^o\\ \\ A=\frac{50\pi}{9}\\ \\ \text{Then we must add the area of the shaded triangle.}\\ \\ A=\frac{bh}{2}\\ \\ \frac{b}{2}=5sin40^o\\ \\ h=5cos40^o\\ \\ A=25sin40^o(cos40^o)\\ \\ \text{Total area is}\\ \\ A=\frac{50\pi }{9}+25sin40^o(cos40^o)\\ \\ A\approx 29.8cm^2$